| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle inside smooth hollow cylinder |
| Difficulty | Standard +0.3 This is a standard circular motion problem requiring resolution of forces in two directions and application of F=mrω². While it involves 3D geometry with a cone and two contact forces, the setup is clearly defined and the solution follows routine mechanics procedures. The constraint that forces are equal (part i) or contact is lost (part ii) makes the algebra straightforward. Slightly above average due to the 3D visualization and two-part structure, but well within typical M2 scope. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F + F\cos60 = mg\) | M1 | Resolves vertically for S |
| \(F = 10m/1.5\) | A1 | May be implied by later work |
| \(F\sin60 = mv^2/0.6\) | M1 | \(10m/1.5 = mv^2/0.6\) |
| \(v = 1.86\text{ ms}^{-1}\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F\cos60 = 10m\) | B1 | May be implied by later work |
| \(F\sin60 = m\omega^2/0.6\) | M1 | |
| \(\omega = 5.37\text{ rads}^{-1}\) | A1 [3] |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F + F\cos60 = mg$ | M1 | Resolves vertically for S |
| $F = 10m/1.5$ | A1 | May be implied by later work |
| $F\sin60 = mv^2/0.6$ | M1 | $10m/1.5 = mv^2/0.6$ |
| $v = 1.86\text{ ms}^{-1}$ | A1 [4] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F\cos60 = 10m$ | B1 | May be implied by later work |
| $F\sin60 = m\omega^2/0.6$ | M1 | |
| $\omega = 5.37\text{ rads}^{-1}$ | A1 [3] | |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6d3892e0-8c88-44ec-940f-c526d71a7fc6-2_268_652_1599_475}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6d3892e0-8c88-44ec-940f-c526d71a7fc6-2_191_323_1653_1347}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A small sphere $S$ of mass $m \mathrm {~kg}$ is moving inside a smooth hollow bowl whose axis is vertical and whose sloping side is inclined at $60 ^ { \circ }$ to the horizontal. $S$ moves with constant speed in a horizontal circle of radius 0.6 m (see Fig. 1). $S$ is in contact with both the plane base and the sloping side of the bowl (see Fig. 2).\\
(i) Given that the magnitudes of the forces exerted on $S$ by the base and sloping side of the bowl are equal, calculate the speed of $S$.\\
(ii) Given instead that $S$ is on the point of losing contact with one of the surfaces, find the angular speed of $S$.
\hfill \mbox{\textit{CAIE M2 2012 Q3 [7]}}